160. Intersection of Two Linked Lists
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Write a program to find the node at which the intersection of two singly linked lists begins.
For example, the following two linked lists:
A: a1 → a2 ↘ c1 → c2 → c3 ↗ B: b1 → b2 → b3
begin to intersect at node c1.
Notes:
- If the two linked lists have no intersection at all, return
null
. - The linked lists must retain their original structure after the function returns.
- You may assume there are no cycles anywhere in the entire linked structure.
- Your code should preferably run in O(n) time and use only O(1) memory.
一道比较经典的问题 求链表交点
如果没有空间复杂度要求的话 可以这样做:
遍历其中一个链表 放入set 遍历第二个 set.contains = true 返回即可
但是要求时间复杂度是O(1) solution比较trick 如下
Two Pointers
- Maintain two pointers pA and pB initialized at the head of A and B, respectively. Then let them both traverse through the lists, one node at a time.
- When pA reaches the end of a list, then redirect it to the head of B (yes, B, that's right.); similarly when pB reaches the end of a list, redirect it the head of A.
- If at any point pA meets pB, then pA/pB is the intersection node.
- To see why the above trick would work, consider the following two lists: A = {1,3,5,7,9,11} and B = {2,4,9,11}, which are intersected at node '9'. Since B.length (=4) < A.length (=6), pB would reach the end of the merged list first, because pB traverses exactly 2 nodes less than pA does. By redirecting pB to head A, and pA to head B, we now ask pB to travel exactly 2 more nodes than pA would. So in the second iteration, they are guaranteed to reach the intersection node at the same time.
- If two lists have intersection, then their last nodes must be the same one. So when pA/pB reaches the end of a list, record the last element of A/B respectively. If the two last elements are not the same one, then the two lists have no intersections.
Complexity Analysis
- Time complexity : O(m+n).
- Space complexity : O(1).
public ListNode getIntersectionNode(ListNode headA, ListNode headB) { //boundary check if(headA == null || headB == null) return null; ListNode a = headA; ListNode b = headB; //if a & b have different len, then we will stop the loop after second iteration while( a != b){ //for the end of first iteration, we just reset the pointer to the head of another linkedlist a = a == null? headB : a.next; b = b == null? headA : b.next; } return a;}
a,b链表的遍历过程会变为这样:
a1->a2->c1->c2->c3->b1->b2->b3->c1->c2->c3
b1->b2->b3->c1->c2->c3->a1->a2->c1->c2->c3
在第二个c1处汇合
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